I have a simple question about notation w.r.t the norms specifcally: $\|g - g_{n}\|_{\infty}$
Question:
i) $\|g - g_{n}\|_{\infty}$: With regards to this question, we showed that the set $A = \{f \in C([0,1]): f(0) = 1 \}$ is closed in the normed space $(C([0,1]), \|f\|_{\infty} = \sup_{x\in [0,1]}|f|)$.
To do this we go through the following proof:
Take a sequence of functions $g_n \in A$ s.t. $\|g_n - g \|_{\infty} \rightarrow 0$ and $g \in C([0,1])$. We want to show $g \in A$.
We study $g(0)$: $$|g(0) - 1| \leq |g(0) - g_{n}(0)| + |g_{n}(0) - 1| \leq \|g - g_{n}\|_{\infty} = \sup_{x \in [0,1]}|g(x) - g_n(x)|$$
So my problem is with the last inequality. Apparently this last inequality: $\|g - g_{n}\|_{\infty} = \sup_{x \in [0,1]}|g(x) - g_n(x)|$ goes to zero to establish convergence. My question is why does this last inequality go to zero? If it is because we are assuming $\|g_n - g \|_{\infty} \rightarrow 0$ in $A$, then why are we allowed to make this assumption about the set?
Recall that in a metric space, to show that a set $A$ is closed you need to show that it contains all of its limit points, i.e. you need to show
"If $x_n \in A$ satisfies $x_n \rightarrow x$, then $x \in A$"
Don't feel uneasy about this. I think what may be confusing here is that it's not immediately clear why $g$, the limit, is continuous. This is a non-trivial (but standard) argument, that you have no doubt seen before. It is an "$\frac{\varepsilon}{3}$ argument". The fact that the limit is continuous is what's called completeness of $C[0,1]$
Your second question about why the last term goes to zero I'll answer first.
$||g_n - g||_{\infty} \rightarrow 0$ means by definition that $\lim_{n\rightarrow \infty}\sup_{x\in[0,1]}|g(x) - g_n(x)| = 0$.
Your proof is essentially fine. But you should carefully justify (even it's just to yourself) each step.
Suppose that $g_n \in A \rightarrow g$ in the $|.|_{\infty}$ norm. Since $C[0,1]$ is complete with respect to this norm, it follows that $g \in C[0,1]$. The condition $g_n \rightarrow g$ implies that $\lim_{n\to\infty}g_n(x) = g(x)$ for all $x \in [0,1]$ (note, this is strictly weaker, but is true because the limit of the supremums goes to zero). In particular, when $x =0$, $g(0) = \lim_{n\to\infty}g_n(0) = \lim_{n\to\infty} 1 = 1$, hence $g \in A$.